Theorem
Find two non-isomorphic groups \(H_1\) and \(H_2\) such that \(\operatorname{Aut}(H_1)\) is isomorphic to \(\operatorname{Aut}(H_2)\).
Proof
Step 1
Let us choose the groups
Let \(H_1 = \mathbb{Z}_3\) and \(H_2 = \mathbb{Z}_4\). Since \(|\mathbb{Z}_3| = 3\) and \(|\mathbb{Z}_4| = 4\), these groups are not isomorphic.
Step 2
Let us compute their automorphism groups
For any cyclic group \(\mathbb{Z}_n\), every automorphism is determined by its action on a generator. Hence,
\[
\operatorname{Aut}(\mathbb{Z}_n, +) \cong U_n,
\]
where
\[
U_n = \{a \in \mathbb{Z}_n \mid \gcd(a, n) = 1\}
\]
is the group of units modulo \(n\).
Step 3
To evaluate \(\operatorname{Aut}(H_1)\) and \(\operatorname{Aut}(H_2)\)
For \(H_1 = \mathbb{Z}_3\):
\begin{align}
\operatorname{Aut}(\mathbb{Z}_3) &\cong U_3 \nonumber\\
&\cong \{1,2\} \nonumber\\
&\cong \mathbb{Z}_2\nonumber
\end{align}
For \(H_2 = \mathbb{Z}_4\):
\begin{align}
\operatorname{Aut}(\mathbb{Z}_4) &\cong U_4\nonumber\\
&\cong \{1,3\}\nonumber\\
&\cong \mathbb{Z}_2\nonumber\\
\end{align}
Step 4
Conclude the Isomorphism of Automorphism Groups
Since both \(\operatorname{Aut}(\mathbb{Z}_3)\) and \(\operatorname{Aut}(\mathbb{Z}_4)\) are isomorphic to \(\mathbb{Z}_2\), it follows that:
\[
\operatorname{Aut}(H_1) \cong \operatorname{Aut}(H_2).
\]